Ascorbic Acid Vitamin C I Contains C , H , and O

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Ascorbic Acid ( Vitamin C ) - I Contains C , H ,
and O

• Upon combustion in excess oxygen, a 6.49 mg
  sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate its Empirical formula!
• C 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
• 2.65 x 10-3 g C
• H 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
• 2.92 x 10-4 g H
• Mass Oxygen 6.49 mg - 2.65 mg - 0.30 mg
• 3.54 mg O

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Vitamin C Combustion - II

• C 2.65 x 10-3 g C / ( 12.01 g C / mol C )
• 2.21 x 10-4 mol C
• H 0.295 x 10-3 g H / ( 1.008 g H / mol H )
• 2.92 x 10-4 mol H
• O 3.54 x 10-3 g O / ( 16.00 g O / mol O )
• 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C 1.00 Multiply each by 3 3.00 3.0
• H 1.32
  3.96 4.0
• O 1.00
  3.00 3.0

C3H4O3
3

Vitamin C
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Determining a Chemical Formula from
Combustion Analysis - I
Problem Erthrose (M 120 g/mol) is an
important chemical compound as
a starting material in chemical synthesis, and
contains Carbon Hydrogen, and
Oxygen. Combustion analysis of
a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Plan We find the masses
of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of
Carbon and Hydrogen are subtracted from
the sample mass to get the mass of
Oxygen. We then calculate moles, and construct
the empirical formula, and from the
given molar mass we can calculate the
molecular formula.
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Determining a Chemical Formula from Combustion
Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2

0.2729 g C
/ 1 g CO2 Mass fraction of H in H2O


0.1119 g H / 1 g H2O Calculating masses of C
and H Mass of Element mass of compound x
mass fraction of element
mol C x M of C mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C 44.01 g
CO2
mol H x M of H mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H 18.02
g H2O
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Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C 1 g CO2
Mass (g) of C 1.027 g CO2 x
0.2803 g C Mass (g) of H 0.4194 g H2O x
0.04693 g H Calculating
the mass of O Mass (g) of O Sample mass -(
mass of C mass of H )
0.700 g - 0.2803 g C - 0.04693 g H 0.37277 g
O Calculating moles of each element C
0.2803 g C / 12.01 g C/ mol C 0.02334 mol C
H 0.04693 g H / 1.008 g H / mol H 0.04656 mol
H O 0.37277 g O / 16.00 g O / mol O
0.02330 mol O C0.02334H0.04656O0.02330 CH2O
formula weight 30 g / formula 120 g /mol / 30 g
/ formula 4 formula units / cpd C4H8O4
0.1119 g H 1 g H2O
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Some Compounds with Empirical Formula CH2O
(Composition by Mass 40.0 C, 6.71 H, 53.3O)
Molecular M Formula
(g/mol) Name Use or Function
CH2O 30.03 Formaldehyde
Disinfectant Biological

preservative C2H4O2 60.05
Acetic acid Acetate polymers vinegar

( 5 solution) C3H6O3
90.08 Lactic acid Causes milk
to sour forms
in muscle
during exercise C4H8O4 120.10
Erythrose Forms during sugar

metabolism C5H10O5
150.13 Ribose Component of
many nucleic
acids and
vitamin B2 C6H12O6 180.16
Glucose Major nutrient for energy

in cells
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Two Compounds with Molecular Formula C2H6O
Property Ethanol
Dimethyl Ether
M (g/mol) 46.07
46.07 Color
Colorless
Colorless Melting point - 117oC
- 138.5oC Boiling point
78.5oC -
25oC Density (at 20oC) 0.789 g/mL
0.00195 g/mL Use
Intoxicant in In
refrigeration
alcoholic beverages
H H
H H H C C O
H H C O C H
H
H H
H
Table 3.4
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Molecular Formula
Molecules
Atoms
Avogadros Number
6.022 x 1023
Moles
Moles
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Chemical Equations
Qualitative Information
Reactants
Products
States of Matter (s) solid (l)
liquid (g) gaseous (aq) aqueous
2 H2 (g) O2 (g) 2 H2O (g)
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Chemical Equation Calculation - I
Atoms (Molecules)
Avogadros Number
6.02 x 1023
Molecules
Reactants
Products
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Chemical Equation Calculation - II
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
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• Stoichiometry
• The calculation of the quantities of reactants
  and products involved in a chemical reaction
• Interpreting a Chemical Equation
• The coefficients of the balanced chemical
  equation may be interpreted in terms of either
  (1) numbers of molecules (or ions or formula
  units) or (2) numbers of moles, depending on your
  needs.

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Information Contained in a Balanced Equation
Viewed in Reactants
Products terms of 2 C2H6 (g)
7 O2 (g) 4 CO2 (g) 6 H2O(g) Energy
Molecules 2 molecules of C2H6 7 molecules of
O2
4 molecules of CO2 6 molecules of
H2O Amount (mol) 2 mol C2H6 7 mol O2
4 mol CO2 6 mol H2O Mass (amu) 60.14 amu
C2H6 224.00 amu O2
176.04 amu CO2
108.10 amu H2O Mass (g) 60.14 g C2H6
224.00 g O2 176.04 g CO2 108.10 g H2O Total
Mass (g) 284.14g
284.14g
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• To find the amount of B (one reactant or product)
  given the amount of A (another reactant or
  product)

• 1. Convert grams of A to moles of A
• ? Using the molar mass of A
• 2. Convert moles of A to moles of B
• ? Using the coefficients of the balanced chemical
  equation
• 3. Convert moles of B to grams of B
• ? Using the molar mass of B

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• Propane, C3H8, is normally a gas, but it is sold
  as a fuel compressed as a liquid in steel
  cylinders. The gas burns according to the
  following equation
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• How many grams of CO2 are produced when 20.0 g of
  propane is burned?

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Molar masses C3H8 3(12.01) 8(1.008) 44.094
g CO2 1(12.01) 2(16.00) 44.01 g
59.9 g CO2 (3 significant figures)
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• Limiting Reactant
• The reactant that is entirely consumed when a
  reaction goes to completion
• Once one reactant has been completely consumed,
  the reaction stops.
• Any problem giving the starting amount for more
  than one reactant is a limiting reactant problem.

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Figure 3.14 Limiting reactant analogy using
cheese sandwiches.
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• All amounts produced and reacted are determined
  by the limiting reactant.
• How can we determine the limiting reactant?
• Use each given amount to calculate the amount of
  product produced.
• The limiting reactant will produce the lesser or
  least amount of product.

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• Magnesium metal is used to prepare zirconium
  metal, which is used to make the container for
  nuclear fuel (the nuclear fuel rods)
• ZrCl4(g) 2Mg(s) ? 2MgCl2(s) Zr(s)
• How many moles of zirconium metal can be produced
  from a reaction mixture containing 0.20 mol ZrCl4
  and 0.50 mol Mg?

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ZrCl4 is the limiting reactant. 0.20 mol Zr will
be produced.
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• Urea, CH4N2O, is used as a nitrogen fertilizer.
  It is manufactured from ammonia and carbon
  dioxide at high pressure and high temperature
• 2NH3 CO2(g) ? CH4N2O H2O
• In a laboratory experiment, 10.0 g NH3 and 10.0 g
  CO2 were added to a reaction vessel. What is the
  maximum quantity (in grams) of urea that can be
  obtained? How many grams of the excess reactant
  are left at the end of the reactions?

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Molar masses NH3 1(14.01) 3(1.008) 17.02
g CO2 1(12.01) 2(16.00) 44.01
g CH4N2O 1(12.01) 4(1.008) 2(14.01)
1(16.00) 60.06 g
CO2 is the limiting reactant. 13.6 g CH4N2O will
be produced.
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To find the excess NH3, we find how much NH3
reacted
Now subtract the amount reacted from the starting
amount
10.0 at start -7.73 reacted 2.27 g remains
2.3 g NH3 is left unreacted. (1 decimal place)
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• Theoretical Yield
• The maximum amount of product that can be
  obtained by a reaction from given amounts of
  reactants. This is a calculated amount.

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• Actual Yield
• The amount of product that is actually obtained.
  This is a measured amount.
• Percentage Yield

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• 2NH3 CO2(g) ? CH4N2O H2O
• When 10.0 g NH3 and 10.0 g CO2 are added to a
  reaction vessel, the limiting reactant is CO2.
  The theoretical yield is 13.6 of urea. When this
  reaction was carried out, 9.3 g of urea was
  obtained. What is the percent yield?

Theoretical yield 13.6 g Actual yield 9.3 g
68 yield (2 significant figures)
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Figure 4.1 Reaction of potassium iodide solution
and lead (II) nitrate solution. Photo courtesy of
James Scherer.
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Figure 4.2 Motion of ions in solution.
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Figure 4.3 Testing the electrical conductivity
of a solution water.Photo courtesy of American
Color.
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Figure 4.3 Testing the electrical conductivity
of a solution sodium chloride.Photo courtesy of
American Color.
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Figure 4.4 Comparing strong and weak
electrolytes HCl. Photo courtesy of American
Color.
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Figure 4.4 Comparing strong and weak
electrolytes NH3. Photo courtesy of American
Color.
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Methanol
Li
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The Role of Water as a Solvent The Solubility
of Ionic Compounds
Electrical conductivity - The flow of electrical
current in a solution is a
measure of the solubility of ionic
compounds or a
measurement of the presence of ions in
solution. Electrolyte - A substance that
conducts a current when dissolved in
water. Soluble ionic compound
dissociate completely and
may conduct a large current, and are called
strong Eeectrolytes.
NaCl(s) H2O(l)
Na(aq) Cl -(aq)
When sodium chloride dissolves into water the
ions become solvated, and are surrounded by water
molecules. These ions are called aqueous and
are free to move through out the solution, and
are conducting electricity, or helping electrons
to move through out the solution
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Fig. 4.2
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The Solubility of Ionic Compounds in Water
The solubility of ionic compounds in water
depends upon the relative strengths of the
electrostatic forces between ions in the ionic
compound and the attractive forces between the
ions and water molecules in the solvent. There
is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called
insoluble compounds may be several orders of
magnitude less than ones that are
called soluble in water, for example
Solubility of NaCl in water at 20oC 365
g/L Solubility of MgCl2 in water at 20oC 542.5
g/L Solubility of AlCl3 in water at 20oC 699
g/L Solubility of PbCl2 in water at 20oC 9.9
g/L Solubility of AgCl in water at 20oC 0.009
g/L Solubility of CuCl in water at 20oC 0.0062
g/L
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Precipitation Reactions Will a Precipitate Form?
If we add a solution containing potassium
chloride to a solution containing ammonium
nitrate, will we get a precipitate?
KCl(aq) NH4NO3 (aq) K(aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have potassium chloride and ammonium
nitrate, or potassium nitrate and
ammonium chloride. In looking at the solubility
table it shows all possible products as soluble,
so there is no net reaction!
KCl(aq) NH4NO3 (aq) No Reaction!
If we mix a solution of sodium sulfate with a
solution of barium nitrate, will we get a
precipitate? From the solubility table it shows
that barium sulfate is insoluble, therefore we
will get a precipitate!
Na2SO4 (aq) Ba(NO3)2 (aq)
BaSO4 (s) 2 NaNO3 (aq)
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